If $b^2>4ac,\text{ then }a{\left(x^2+4x+4\right)}^2+b\left(x^2+4x+4\right)+c=0$ has distinct roots if :

Putting x² + 4x + 4 =(x +2)2 = y.
We have ay² + by + c = 0. ...(i)
b² - 4ac > 0 $\Rightarrow$ equation (i) has real and distinct roots.
Also $y=(x+2{)}^2>0$ as all four roots are distinct
(i) has both roots > 0
sum of the roots $=-\frac{b}{a}>0$
product of roots $=\frac{c}{a}>0$
o and b arc of opposite sign and a and c are of same sign. Only choice (c) satisfies these conditions.